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Prove that, if $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

Proposition: If $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$, then there is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

A (hypothesis): $a$ and $b$ are real numbers, at least one of which is not $0$, and $i = \sqrt{−1}$.

B (conclusion): There is a unique complex number, say $c + di$, such that $(a + bi)(c + di) = 1$.

My Work

B1: $(a + bi)(c + di) = 1$

$\therefore a + bi \not = 0$

B2: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

A1: $c + di = \dfrac{1}{a + bi}$ where $a + bi \not = 0$

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$.

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A2: There is a complex number $x + yi$ such that $x + yi = \dfrac{1}{a + bi}$.

$\implies (a + bi)(x + yi) = 1$ where $a + bi \not = 0$

A3: $(a+bi)(x + yi) = (a + bi)(c + di)$ where $a + bi \not = 0$

$\implies x + yi = c + di$

$Q.E.D.$

EDIT: My Work #2

A1: $(a + bi)(c + di) = 1$

$\implies ac + adi + bic + bdi^2 = 1$

$\implies ac + adi + bic - bd = 1$

$\implies (ac - bd) + i(ad + bc) = 1$ is a complex number, where $(ac - bd)$ is the real part and $i(ad + bc)$ is the imaginary part.

A2: Let $ac - bd = 1$ and $ad + bc = 0$

$\therefore 1 + 0 = 1$

Therefore, in A1, I have constructed the object specified in the conclusion ($c + di$).

Now, below, I use the direct uniqueness method to show that the object is unique.

A3: Let $(a + bi)(x + yi) = 1$

A4: $(a + bi)(c + di) = (a + bi)(x + yi)$ where ($a \lor b) \not = 0$.

$\implies c + di = x + yi$

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback. If there are any errors, please explain why and what the correct procedure is.

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    You've only renamed $\frac{1}{a+bi}$ into $x+yi$ and from there concluded that $\frac1{a+bi}$ exists, when what you're proving in the first place is that what we intuitively mean by $\frac{1}{a+bi}$ is something that exists and makes sense. That's a bit backwards. Instead, try to find out what $c$ and $d$ are, specificallly, using the assumption that $a+bi \neq 0$. – 2017-02-19
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    What is your definition of "complex number"? Do you already know that division of complex numbers is defined? – 2017-02-19
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    This question is rather strange: at no moment, the word "inverse" has been used as if it is a dirty word... – 2017-02-19
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    If you're allowed to use the fact that modulus-argument form exists and is unique, then finding $c$ and $d$ is not necessary, because it's easy to invert a number in modulus-argument form. – 2017-02-19

3 Answers 3

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The aim of this exercise is to prove that the thing that you denote by $\frac1{a+bi}$ uniquely exists. Using this notation a priori implicitly assumes the statement we are about to prove.

Hint: Instead, as commented, try to explicitly find $c$ and $d$. For this, observe that $(a+bi)(a-bi)$ is real, and division is already known among reals.

For uniqueness of such $c+di$, we can take a general argument:
Suppose $w_1$ and $w_2$ are both inverses of $z$, i.e. $w_1z=zw_2=1$. Then we have $$w_1 = w_1\cdot 1 = w_1zw_2=1\cdot w_2=w_2$$

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    Thanks for the response. Can you please explicitly explain how to construct the object $c + di$? I don't understand why finding $c$ and $d$ helps in constructing $c + di$ . – 2017-02-19
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    @ThePointer Because that tells you that given $a$ and $b$ we can find $c$ and $d$ such that $c+di$ behaves the way we would expect $\frac1{a+bi}$ to do. This, again, means that you can _assign_ $\frac1{a+bi}$ to specifically mean $c+di$. Until then, you can make intuitive sense of the concept of $\frac1{a+bi}$, because you have experience with fractions already, but you have no guarantee that it exists, or if it exists, _when_ it exists or whether it's unique. – 2017-02-19
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    @ThePointer See my answer. – 2017-02-19
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    @TakahiroWaki thanks for the response, but your answer does not tell me anything useful: There is no explanation at all. – 2017-02-19
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    @ThePointer I showed $\dfrac{1}{a+bi}=c+di$. – 2017-02-19
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    I have edited the OP with a second solution. I would greatly appreciate it if someone can provide feedback on whether this is correct. – 2017-02-19
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I will start by trying to answer this explicit part of your question.

I would greatly appreciate it if people could please take the time to review my proof for correctness and provide feedback. If there are any errors, please explain why and what the correct procedure is.

In your first attempt you begin with the statement $$ (a+bi)(c+di) = 1 . $$ Since all you know from the hypothesis is the number $a+bi$ that beginning statement makes no sense. There is no $c+di$ in the hypothesis. No correct proof could start that way.

About your A1 you say

Therefore, in A1, I have constructed the object specified in the conclusion (c+di).

but that's not true. You have just written "$1/(a+bi) = x+yi$". To construct it you have to tell us the values of $x$ and $y$ in terms of $a$ and $b$.

Where should you start? I suggest thinking about an example. Suppose you begin with $a+bi = 3+4i$. Then what you must do is find some (as yet unknown) complex number $c+di$ so that $$ (3+4i)(c+di) = 1 . $$

Well the second line in your A1 "proof" says you know what to do next: $$ (3+4i)(c+di) = (3c -4d) + (3d + 4c)i = 1 . $$ That tells you $$ \begin{align} 3c - 4d & = 1 \\ 3d + 4c & = 0 \end{align} $$ Now you can solve those two simultaneous equations in two unknowns and show that they have a unique solution.

Once you understand the example $3+4i$, you can do the same thing algebraically for the general complex number $a+bi$.

Hint. Your answer will have a square root in it. In the example that square root turned out to be an integer.

Note. This is a pretty longwinded way to go about the proof. There are cleverer shorter ways. But this is how you could approach the problem without knowing about the clever stuff.

Edit to answer the OP's last question.

There are several ways to show that the inverse $c+di$ is unique. One is to observe that the equations above you solved to find it have a unique solution. Another is to note that for any three complex numbers $\alpha$, $\beta$ and $\gamma$ with $\alpha \ne 0$, if $$ \alpha \beta = \alpha \gamma $$ then $$ \alpha (\beta - \gamma) = 0 $$ so $$ \beta - \gamma = 0 $$ (not because you "divide by $\alpha$" - because you check the arithmetic) hence $$ \beta = \gamma. $$

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    **A** represents working forwards from the hypothesis, whilst **B** represents working backwards from the conclusion. – 2017-02-19
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    See my edits. Please study this answer and @Berci 's before you post more questions in comments. – 2017-02-19
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    I got $d = \dfrac{-b}{b^2 + a^2}$ and $c = \dfrac{a}{b^2 + a^2}$. Therefore, since $(c + di)(a+bi) = 1$, we should have $\dfrac{a}{b^2 + a^2} * \left( \dfrac{-b}{b^2 + a^2} \right)i = 1$, right? But when I expand $\dfrac{a}{b^2 + a^2} * \left( \dfrac{-b}{b^2 + a^2} \right)i$, it seems that I am not getting $1$? – 2017-02-19
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    Your $d$ and $c$ are right. Your check is wrong. You haven't done the complex multiplication correctly. – 2017-02-19
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    sorry, I wrote incorrectly. I meant that, for that $c + di$, I am not getting $(c + di)(a + bi) = 1$. – 2017-02-19
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    With the values of $c$ and $d$ you found you _will_ get that $(a+bi)(c+di) = 1$ - if you do the complex number arithmetic correctly. Try it with $a+bi = 3+4i$. – 2017-02-19
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    Ok, I understand the construction of the object. Isn't my proof of uniqueness correct? Also, I don't think what you said about using $c + di$ in the beginning is correct. It isn't part of the hypothesis but that doesn't mean we can't create it to start our proof -- even you did that! I don't see how it would be possible to construct a proof without first constructing the object $c + di$ in the beginning of our proof, despite it not being in the hypothesis. Notice that you did the same thing. – 2017-02-19
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    Yes you start the proof by constructing it. But that's not what you did in your question. – 2017-02-19
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    For uniqueness: **A6:** Let $(x + yi)(a + bi) = 1$. **A7:** Therefore, we have $(x + yi)(a + bi) = 1$ and $(c + di)(a + bi) = 1 \implies (x + yi)(a + bi) = (c + di)(a + bi) \implies x + yi = c + di$ by dividing both sides by $(a + bi)$. Is this correct for showing uniqueness? – 2017-02-20
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    How do you know you can "divide by $a+bi$"? Proving that's OK Is the whole point of the exercise. See my edit. You're on your own here from now on. – 2017-02-20
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It seem that both proofs are strange or incomplete.

$\dfrac1{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i$

This is a unique complex number.

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    But you haven't proven that it's the only answer. – 2017-02-19
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    @Arthur "please take the time to review my proof for correctness" . – 2017-02-19